A reply to “The Extinction of Pi: The short version”

We’re very proud to have a guest author today. If you’ve been following the comments on Ex Falso… you will have noticed Dan, who kindly offered to write replies to Mathis’ more math-based papers. He took a look at Mathis’ ‘proof’ that (in kinematic situations) and promptly sent us this article. We’re happy to have Dan on board and are looking forward to more articles!

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In The Extinction of Pi: the short version Mathis claims to have a simple geometric proof that Pi=4.

Before I begin I feel I should comment on Mathis’ method of proof on this one. This is one of the rare occasions where his ideas are easy to understand as there is no excess verbiage, no ill defined terms, only one point where the verbal explanation does not quite match the diagram (and an honest mistake at that), and no appeals to unrelated or ambiguous notions like ‘velocity of the radius’, although you may find this happens more frequently in long version of The Extinction of Pi. Many mathematicians have different requirements before they will call a proof elegant. If all known proofs of a statement require long and intricate arguments or “heavy machinery” and a new proof which relies solely on elementary geometry is found, then they will for the most part agree that it is elegant. All things considered, had his conclusions been accurate or his method sound, then this is the only time I have seen Mathis present what would be called an elegant proof.

Disproof #1

Left: Mathis diagram --- Right: my diagram

To quote Mathis himself:

Given the diagram above, which is part of a circle, we let

AO = AB

This makes the angle at B equal to 45°. Which means that

DB = DC.

This must mean that, by substitution,

AD + DC = AB

Now, the distance AD + DC can be redrawn as any number of sub-distances, as above.

Here we divide the distance into four sub-distances, taking us from A to C in 4 steps.

This new path, drawn by the sub-distances, is equivalent in length to AD + DC, as can be seen by any cursory analysis. It is nearer the path of the arc AC, but it retains the original distance of AD + DC.

If we divide the distance into eight sub-distances instead of four, we approach the path of the arc AC even nearer; but, again, we retain the original distance of AD + DC.

If we take this process to its limit, we take our path to the path of the arc AC.

This must mean that the length of the arc AC is equal to the distance AD + DC.

By substitution, this means that

arcAC = AB

Since 8 such arcs make up the circle, the circumference of the circle is 8AB.

Since AB = AO, and AO is the radius, we have found that

C = 8r

π does not exist in the circle equation. It is extinct.

The above is Mathis’s “proof” that Pi = 4, at first glance it seems plausible, the little ‘steps’ do visually look more and more like the arc of the circle as they are made ever smaller. All computer graphics are done in the same spirit; the pixels an LCD monitors are tiny squares, so any time anything curved is displayed on an LCD monitor it is being approximated visually by these squares. However a visual approximation of a curve is not the same as a curve. The error I mentioned where his explanation does not match the diagram involves his letter C. He uses it once as a point on the graph and then later as the circumference of the circle. This is not a major mistake and does not in any way change his argument or my proof that his method is flawed. I will now refer to Mathis’ method as the ‘step method’ because when drawn; the approximation looks like steps in a staircase. My proof that his method is invalid will now follow.

Definition and Labels

1. Let a curve be the locus of points satisfying the equation y = f(x) where f is a continuous function or equivalently the set {(x,y) : xεR, y=f(x), f is continuous}. (think about drawing a curve above the x-axis where you never have to lift your pencil and above each x there is only one point on the curve and it never intersects itself)
2. Let l(Z) denote the length of the curve Z
3. In my diagram above, let R and B denote the red curve from (0,0) to (1,1) and the straight blue line from (0,0) to (1,1) respectively.
4. A curve representing the function f(x) is said to be monotonic and strictly increasing if x<y implies f(x) < f(y) (If y is the right of x, then the point on the curve above y is higher than the point on the curve above x)
5. A curve representing the function f(x) is said to be monotonic and strictly decreasing if x f(y) (If y is the right of x, then the point on the curve above y is lower than the point on the curve above x)
6. A straight line is the curve between two points with the shortest distance. So if A is a straight line connecting any two points, and B is any curve distinct from A connecting the same two points we have l(A) < l(B) where l(Z) is the function that gives the length of the curve.

Assume that Mathis is correct and the length of a curve can be approximated by considering the lengths of the horizontal and vertical line segments that make up the ‘steps’ in the diagrams. Mathis is correct, no matter how few or how many of these steps we subdivide the interval [0,1] into, the total of their lengths will remain constant. They steps can be made small, medium, large, evenly sized, or unevenly sized yet their total length will remain the same. The below equation demonstrates this for evenly sized steps, which I will point out can only be drawn when the curve you are approximating is an arc of a circle or a straight line.

Now consider the curve R and B in the right side of the diagram. They can be approximated by ‘steps’ as in the Mathis proof. Notice that the big triangle in black has two sides, each of length 1 giving a combined length of 2. Thus the total length of the steps approximating B must be 2 implying the length of B is 2. The steps that approximate R must have total length 2 implying R has total length 2. We have shown that l(B) = l(R) or that the length of B and the length of R are in fact equal. However by assumption since B is a straight line it has the shortest possible length, and since R is a curve distinct from B it must have length greater than B. We have two contradictory statements: l(B) = l(R) and l(B) < l(R). The assumption that the ‘step method’ of calculating length is accurate leads to a contradiction and is therefore invalid.
QED

One can extend the above line of reasoning to show a result of the Mathis method is the absurd idea that if m and n are any two distinct points, and X & Y are two distinct monotonic strictly increasing/decreasing curves joining m to n, then X and Y must have the same length. This would mean the length of the curves defined by the following functions are identical on [0,1]: f(x) = sin(Pi*x/2), g(x) = x, h(x) = x^2, j(x) = x^3, or even k(x) = x^n for any integer n>0.

Disproof #2

Assume that the Mathis’ “step method” can be used to calculate the length of a curve. In this situation our curve is the segment AC.
Let ABC be a right triangle with side lengths |AB| = 1, |AC| = 2, and |BC| unknown.
If we accept Mathis’ “step method” then we can approximate line segment AC by the red steps, and approximate the length |AC| by the lengths of the steps. As we make the steps smaller and smaller and pass to the limit, we have the length of the segment equal to the limit of the sum of the lengths of the steps, so as Mathis has demonstrated we have |AC| = |AB| + |BC|. Plugging in our numbers for |AC| and |AB| we get the equation 3 = 1 + |BC| which gives |BC| = 2.
From basic trigonometry we have sin(angle BAC) = opposite/hypotenuse = |BC|/|AC| = 2/2 = 1.

Now again from basic properties of triangles we can deduce that triangle ABC is “half” of an equilateral triangle. That is, if we reflected the triangle along the line segment BC and let A’ be the mirror image of A, then triangle ACA’ would be equilateral, as |AA’| = |AC| = |CA’|. This allows us to calculate angle BAC to be Pi/3 radians or 60 degrees.
Now given that we have found angle BAC to be Pi/3 radians we can calculate the sin of that angle in radians as sin(Pi/3) = sqrt(3)/2, or sin(60 degrees) = sqrt(3)/2)
So we have calculated sin(angle BAC) two different ways, and they should be the same.
Equating sin(BAC) = sin(BAC) and doing a few simple manipulations:

sin(BAC) = sin(BAC)
1=sqrt(3)/2
1^2 = [sqrt(3)/2]^2
1 = 3/4
4 = 3
4 – 3 = 3 – 3
1 = 0

Thus the assumption that the Mathis “step method” is an accurate way to calculate the length of a curve leads to the absurd conclusion that 1 = 0.
QED

Afterthoughts

When I first set out to write this out I wasn’t entirely sure how to go about it. I found myself going on lengthy rants about Mathis’ ideas, methods, and conclusions. I could write a long piece attacking some of his ideas and explaining the folly in his way of thinking, but I had set out to discredit one specific paper. So I edited out my own rants and tried to make this as straight forward as possible and attack the method, not the man. I however must point out that the number Pi is so entrenched and intricately related to so many areas of mathematics and real world applications that if the value of Pi were to change, then everything in the following list would fall apart or at the very least need a significant rethink. So by Mathis attacking the value of Pi he is indirectly attacking so much more.
Here is a VERY short list of subjects and real world applications that Pi plays role in (it is more ‘important’ in some area than others):

• complex analysis (contour integration in the complex plane)
• probability, simulation, and testing
• plane geometry
• the GPS navigation system
• signal processing (TV, radio, satellite)
• physics (it shows up everywhere!)
• real analysis (estimation see Stirling’s formula)

In closing I would like to quote Mathis from his long version of the Extinction of Pi, make a few short observations, and leave a question for Mathis and another for his supporters (I would really love to read a response!)

I show that in kinematic situations, Pi is 4. For all those going ballistic over my title, I repeat and stress that this paper applies to kinematic situations, not to static situations. I am analyzing an orbit, which is caused by motion and includes the time variable. In that situation, Pi becomes 4. When measuring your waistline, you are not creating an orbit, and you can keep Pi for that. So quit writing me nasty, uninformed letters.
-Miles Mathis

Pi is a constant so it never changes value based on the situation you are using it. I can only assume what he means is that in a kinematic situation he feels the constant Pi = 3.14159265… is an inappropriate constant to use and the number 4 should be used in its place. I promise you that that is not that case, but a proof of that will have to come at another time. I also acknowledge that I have not proved that Pi = 3.14159265, rather I have proved his method is flawed and any results derived from it are meaningless. There are many algorithms to calculate the exact value of Pi, one involves summation of the power series representation of the tangent function, some involve continued fractions, and some involve summation of other infinite series. Whatever the method used, the tools required are well outside the scope of this paper. To be completely honest, if you really believe that Pi equals anything other than the standard decimal representation, then the knowledge needed to understand methods of computing Pi is also well outside your grasp. Of the Mathis supporters I ask: what do you think is more likely; that in the history of scientific thought everyone else has been mistaken and the only person to truly understand the material is Mathis, or rather that Mathis misunderstands the very fundamentals and thus misunderstands anything built upon them? Take a minute to think about this and answer yourself honestly. My question to Mathis is if this notion Pi=4 only applies to kinematic situations, and if we assume by measuring your waistline you mean Pi = 3.14159265… for the purposes of plane geometry, then why did you bother with this little “proof” if you knew it was wrong?

– (by Dan)